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Suppose $A \in \mathbb{R}^{nxn}$ symmetric and positive definit.

We can show that: $$\exists \epsilon > 0 \in \mathbb{R}: \forall x \in \mathbb{R}^n: x^TAx \ge \epsilon \Vert x \Vert^2$$

My claim is that the smallest eigenvalue of $A$ is the supremum for those $\epsilon$.

I'd appreciate to see a prove/disprove!

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2 Answers 2

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Yes, this is true. Take diagonalization $A=TDT^t$ with $D=\operatorname{diag}(\lambda_1,\ldots \lambda_n)$. We obtain with $\lambda_0:=\min_{i=1..n}\lambda_i$ \begin{align} (Tx)^tATx&=x^tT^tATx=x^tDx=\sum_{i=1}^n \lambda_i x_i^2 \geq \lambda_0\sum_{i=1}^n x_i^2 = \lambda_0 \|x\|^2 = \lambda_0 x^tx \\&= \lambda_0 x^tT^tTx= \lambda_0 (Tx)^tTx = \lambda_0\|Tx\|^2. \end{align} By the substitution $x=T^ty$ follows $y^tAy\geq \lambda_0\|y\|^2$. Its clear that this inequality is sharp (take an eigenvector of $\lambda_0$ as y).

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Let's solve the following minimization problem: $$ \begin{aligned} &\min x^TAx\\ &s.t.\ x^Tx=1. \end{aligned} $$ Using the Lagrange method: $$ L=x^TAx-\lambda(x^Tx-1). $$ So $$ \frac{\partial L}{\partial x}=x^T(A+A^T)-2\lambda x^T=2(Ax)^T-2\lambda x^T=0, $$ or $Ax=\lambda x$, and the minimum is reached at an eigenvector $x$. This immediately proves that indeed for any $\epsilon\le\lambda$ and for any $x$ the inequality $x^TAx\ge\epsilon x^Tx$ (and backward, if the inequality holds, then $\epsilon\le\lambda$).

Actually, we never used the fact that $A$ is positive definite. The statement holds for any symmetric matrix, just $\epsilon$ might be negative in that case.

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